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Lesson: Chapter - 10

Evaluating Functions

Evaluating a function simply means finding f(x) at some specific value x. The Math IC will likely ask you to evaluate a function at some particular constant. Take a look at the following example:

If f(x) = x2 – 3, what is f(5)?

Evaluating a function at a constant involves nothing more than substituting the constant into the definition of the function. In this case, substitute 5 for x:

f(5) = 52 - 3 = 22 It’s as simple as that.

Video Lesson - Value of a Function

The test may also ask questions in which you are asked to evaluate a function at a variable rather than a constant. For example:

If f(x) = 3x/ 4–x, what is f(x + 1)?

To solve problems of this sort, follow the same method you used for evaluating a function at a constant: substitute the variable into the equation. To solve the sample question, substitute (x + 1) for x in the definition of the function:

Operations on Functions

Functions can be added, subtracted, multiplied, and divided like any other quantity. There are a few rules that make these operations easier. For any two functions f(x) and g(x):

Video Lesson - Operation on Function

Rule Example
Addition (f + g)(x) = f(x) + g(x) If f(x) = sin x, and g(x) = cos x: (f + g)(x) = sin x + cos x
Subtraction (f - g)(x) = f(x) - g(x) If f(x) = x2 + 5, and g(x) = x2 + 2x + 1: (fg)(x) = x2 + 5 – x2 – 2x – 1 = –2x + 4
Multiplication (f × g)(x) = f(x) × g(x) If f(x) = x, and g(x) = x3 + 8: (f × g)(x) = (x) × ( x3 + 8) = x4 + 8x
Division If f(x) = 2 cos x, and g(x) = 2 sin2 x:

As usual, when dividing, you have to be aware of possible situations in which you inadvertently divide by zero. Since division by zero is not allowed, you should just remember that any time you are dividing functions, like f(x)/g(x), the resulting function is undefined whenever the function in the denominator equals zero.

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